Plz answer questions no 8

Dear Student,

Take L.H.S.

\sqrt{\frac{\cos e c x - c o t x}{\cos e c x + c o t x}} = \sqrt{\frac{\cos e c x - c o t x}{\cos e c x + c o t x} \times \frac{\cos e c x - c o t x}{\cos e c x - c o t x}} = \sqrt{\frac{{(\cos e c x - c o t x)}^{2}}{{(\cos e c x)}^{2} - {(c o t x)}^{2}}} = \cos e c x - c o t x . . . . . . . . . . . . . . . . . . . . . . . . . . . (A s {(\cos e c x)}^{2} - {(c o t x)}^{2} = 1) = \frac{1}{\sin x} - \frac{\cos x}{\sin x} . . . . . . . . . . . . . . . . . . . . . . . . . . . (A s \cos e c x = \frac{1}{\sin x} a n d c o t x = \frac{\cos x}{\sin x}) = \frac{1 - \cos x}{\sin x} = R . H . S . H e n c e p r o v e d P r o o f o f {(\cos e c x)}^{2} - {(c o t x)}^{2} = 1 {(\cos e c x)}^{2} - {(c o t x)}^{2} = 1 \Rightarrow {(\frac{1}{\sin x})}^{2} - {(\frac{\cos x}{\sin x})}^{2} . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . (A s \cos e c x = \frac{1}{\sin x} a n d c o t x = \frac{\cos x}{\sin x}) \Rightarrow \frac{1 - {(\cos x)}^{2}}{{(\sin x)}^{2}} \Rightarrow \frac{{(\sin x)}^{2}}{{(\sin x)}^{2}} = 1

Regards,